Presentation on theme: "Acids and Bases Topic # 14 Brønsted-Lowry Definitions"— Presentation transcript:
1 Acids and Bases Topic # 14 Brønsted-Lowry Definitions The Ion Product for WaterThe pH and Other “p” ScalesAqueous Solutions of Acids and BasesHydrolysisThe Common Ion EffectBuffer SolutionsIndicators and TitrationsPolyprotic Acids
2 Arrhenius definitions Acid Anything that produces hydrogen ions in a water solution.HCl (aq) H+ + Cl-Base Anything that produces hydroxide ions in a water solution.NaOH (aq) Na OH-Arrhenius definitions are limited to aqueous solutions.
3 Brønsted-Lowry definitions Expands the Arrhenius definitionsAcid Proton donorBase Proton acceptorThis definition explains how substances like ammonia can act as bases.NH3(g) + H2O(l) NH OH-
4 Brønsted-Lowry definitions Strong acids and basesconsidered to ionize completely.HCl(aq) + H2O(l) H3O+ (aq) + Cl-(aq)NaOH(aq) + H2O(l) Na+(aq) + OH-(aq)Weak acids and basesdo not ionize completely.HC2H3O2 (aq) + H2O(l) H3O+(aq) +C2H3O2-(aq)NH3 (aq) + H2O(l) NH4+ (aq) + OH- (aq)
5 Brønsted-Lowry definitions When a Brønsted-Lowry acid dissolves in water, the water acts as a base.HC2H3O2 (aq) + 2H2O(l) H3O+(aq) + C2H3O2-(aq)Acid Base Acid BaseH3O+ (aq) is called the hydronium ion.However, the exact number of water moleculesrequired to hydrate a proton is not known.
6 Brønsted-Lowry definitions Conjugate acid-base pairs.Acids and bases that are related by loss or gain of H+ as H3O+ and H2O.Examples. Acid BaseH3O+ H2OHC2H3O2 C2H3O2-NH4+ NH3H2SO4 HSO4-HSO4- SO42-
7 Common acids and bases Acids Formula Molarity* nitric HNO3 16 hydrochloric HClsulfuric H2SO4 18acetic HC2H3O2 18Basesammonia NH3(aq) 15sodium hydroxide NaOH solid*undiluted.
8 Common Acids and bases Acidic Basic Citrus fruits Baking soda Aspirin DetergentsCoca Cola Ammonia cleanersVinegar Tums and RolaidsVitamin C Soap
9 HC2H3O2 (aq) + H2O(l) H3O+(aq) + C2H3O2-(aq) Water is an amphoteric substance that can act either as an acid or a base,HC2H3O2 (aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)acid base acid baseH2O(l) + NH3(aq) NH4+(aq) + OH-(aq)acid base acid base
10 Auto-ionization of water Auto-ionization When water molecules react with each another to form ions.H2O(l) + H2O(l) H3O+(aq) + OH-(aq)(10-7M) (10-7M)Kw = [ H3O+ ] [ OH- ]= 1.0 x at 25oCNote: [H2O] is not included in expressionbecause H2O is a pure liquid!ion productof water
11 Ion product for waterKw is a temperature dependent equilibrium constant. We commonly use 25oC as the standard temperature.Temperature, oC Kwx 10-15x 10-15x 10-14x 10-14x 10-14
12 Autoionization of water [H+] and [OH-] are always present in aqueous solutions. Only for a neutral solution are they at the same concentration.Neutral solution.[H+] = 10-7 M = [OH-]Acidic solution.[H+] > 10-7 M > [OH-]Basic solution.[H+] < 10-7 M < [OH-]
13 pH and other “p” scalesWe need to measure and use acids and bases over a very large concentration range. And many times knowing thoseconcentrations is very important.pH and pOH are systems to keep track of these very large ranges.pH = -log[H3O+]pOH = -log[OH-]pH + pOH = 14
14 pH calculations Determine the following. pH = -log[H+] pH of 6.7 x 10-3 M H+= 2.2pH of 5.2 x M H+= 11.3[H+], if the pH is 4.5= 3.2 x 10-5 M H+
15 pOH examples Determine the following. pOH = -log[OH-] = 14 - pH pOH of 1.7 x 10-4 M NaOHpOH = pH = 10.2pOH of 5.2 x M H+pH = pOH = 2.7[OH-] , if the pH is 4.5pOH = 9.5[OH-] = 3.2 x M
16 pH scaleA logarithmic scale used to keep track of the large changes in [H+].10-14 M M MVery Neutral VeryBasic AcidicWhen you add an acid, the pH gets smaller.When you add a base, the pH gets larger.
17 pH of some common materials Substance pH1 M HClGastric juicesLemon juiceClassic CokeCoffeePure WaterBloodMilk of MagnesiaHousehold ammonia1M NaOH
18 Acid and Base StrengthStrong acids Ionize completely in water. HCl, HBr, HI, HClO3,HNO3, HClO4, H2SO4.Weak acids Partially ionize in water.Most acids are weak.Strong bases Ionize completely in water Strong bases are metal hydroxides - NaOH, KOHWeak bases Partially ionize in water.no Ka !100% Ionization!no Kb !100% Ionization!
19 These bases are considered to be strong. LiOHlithium hydroxideNaOHsodium hydroxideKOHpotassium hydroxideRbOHrubidium hydroxideCsOHcesium hydroxide*Ca(OH)2calcium hydroxide*Sr(OH)2strontium hydroxide*Ba(OH)2barium hydroxide* Completely dissociated in solutions of 0.01 M or less. These are insoluble bases which ionize 100%. The other five in the list can easily make solutions of 1.0 M and are 100% dissociated at that concentration.
20 Acid and Base StrengthFor strong acids and bases, we can directly calculate the pH or pOH if we know the molar concentration.Examples.0.15 M HCl would produce 0.15 M H+.pH = -log(0.15) = 0.82pOH = pH = =0.052 M NaOH produces M OH-.pOH = -log(0.052) = 1.28pH = pOH = =
21 Acid dissociation constant, Ka The ionization of a weak acid can be expressed as an equilibrium.HA (aq) + H2O(l) H3O+(aq) + A- (aq)The strength of a weak acid is related to its equilibrium constant, Ka.Ka =We omit water.It’s a pure liquid.[A-][H3O+][HA]
22 HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq) Weak acid equilibriaExampleDetermine the pH of a 0.10 M benzoic acid solution at 25 oC if Ka = x 10-5HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq)The first step is to write the equilibrium expression.Ka =[H3O+][Bz-][HBz]
23 Weak acid equilibria HBz H3O+ Bz- Initial conc., M 0.10 0.00 0.00 Change, DM x x xEq. Conc., M x x x[H3O+] = [Bz-] = xWe’ll assume that [Bz-] is negligible compared to [HBz]. The contribution of H3O+ from water is also negligible.
24 Weak acid equilibria Solve the equilibrium equation in terms of x Ka = x =x = (6.28 x )(0.10)= MpH = 2.60x20.10
25 Dissociation of bases, Kb The ionization of a weak base can also be expressed as an equilibrium.B (aq) + H2O(l) BH+(aq) +OH- (aq)The strength of a weak base is related to its equilibrium constant, Kb.Kb =[OH-][BH+][B]
26 Ka and Kb values For weak acids and bases Ka and Kb always have values that are smaller than one.Acids with a larger Ka are stronger than ones with a smaller Ka.Bases with a larger Kb are stronger than ones with a smaller Kb.For a conjugate acid:base pair, pKa + pKb = 14 = pKwMost acids and bases are weak.
28 CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) HydrolysisReaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction.CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)This type of reaction is given a special name.HydrolysisThe reaction of an anion with water to produce the conjugate acid and OH-.The reaction of a cation with water to produce the conjugate base and H3O+.
29 Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+(aq) HydrolysisThere is only a small # of cations with a high enoughpositive charge that can act as acids in hydrolysis.Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+(aq)These metal ions are able to pull electrons from the H-OH bond. If the pull is strong enough, water is split.One of the surrounding water molecules will take on the H+ and form H3O+.
30 Common ion effect (Just an example of Le Châtelier’s principle.) The shift in equilibrium caused by the addition of an ion formed from the solute.Common ionAn ion that is produced by more than one solute in an equilibrium system.Adding the salt of a weak acid to a solution of weak acid is an example of this.
31 Common ion effect Example What is the concentration of hydrogen ions in a solution formed by adding mol of sodium acetate to one liter of M acetic acid (HAc). (Ac = C2H3O2-)(assume that the volume of the solution does not change when the salt is added)[Ac-] [H+][HAc]Ka == 1.7 x 10-5
32 Common ion effectHAc Ac- H+Init. Conc., MChange, DM -x +x +xEq. Conc., M x x xTo solve for x, substitute the equilibrium values into the expression.Ka [HAc][Ac-](1.7 x 10-5) (0.099-x)( x)x ==
33 Common ion effectIf we assume that x is negligible compared to and 0.097, we can simplify the problem to:x =x = 1.7 x 10-5 = [H+]Note. When a solution contains equal concentrations of an acid and its conjugate base, [H+] = Ka.(1.7 x 10-5) (0.099)(0.097)
35 Ions of Neutral Salts Salt Hydrolysis A salt is formed between the reaction of an acid and a base.HCl + NaOH NaCl + H2OUsually, a neutral salt is formed when a strong acid and a strong base are neutralized in the reaction.CATIONSNa+K+Rd+Cs+Mg+2Ca+2Sr+2Ba+2ANIONSCl-Br-I-ClO4-BrO4-ClO3-NO3-
36 These ions have little tendency to react with water, they just In this NEUTRAL environment, all the ions are spectator ionsfrom the strong acid and strong base reaction. There are noinsoluble particles.These ions have little tendency to react with water, they juststay dissociated into 100% ions.So, salts consisting of these ions are called neutral saltsand the resulting solution has a pH of 7.For example: NaCl, KNO3, CaBr2, CsClO4 are neutral salts.
37 When weak acids and weak bases react, the strength of the stronger ion conjugate acid or conjugate base in the salt determines the pH of its solutions.Acidic IonsNH4+Al+3Pb+2Sn+2TransitionMetalIonsHSO4-H2PO4-Basic IonsF-C2H3O2-NO2-HCO3-CN-CO3 -2S-2SO4-2HPO4-2PO4-3
38 The resulting salt solution can be acidic, neutral or basic. A salt formed between a strong acid and a weak base is an acid salt, for example NH4Cl:( HCl + NH3 NH4Cl )Strong weakThe NH4Cl is soluble and reacts with H2O:NH H2O NH3 + H3O+The act of the salt particles reacting with water moleculesin the ‘salt solution’ is HYDROLYSIS.
39 A salt formed between a weak acid and a strong base is a basic salt, for example NaCH3COO. (CH3COOH + NaOH NaCH3COO + H2O)CH3COO H2O CH3COOH + OH-pH > 7
40 Deciding whether a salt is acidic, basic or neutral! ******Decide which ion in the salt could producea stronger acid or stronger base.Example:NaF NaOH or HFNaOH is the stronger base soNaF is a basic saltNaClO NaOH or HClO4BOTH are strong so neutral salt!Fe(NO3) Fe(OH)2 or HNO3HNO3 is the stronger compoundFe(NO3)2 is an acidic salt
41 What if both ions are from weak acids and weak bases? Salt: NH4NO NH4OH and HNO2weak base and weak acidCan compare Kb for NH4OH and Ka for HNO2:Kb NH4OH = 1.6 x10-5Ka for HNO2 = 4.5 x 10-4Ka > Kb so the salt is acidic!
42 Ka2 for H2CO3 = 4.8 x 10 -11 So (NH4)2CO3 is basic! Another example: ( qt.1h. on Salt Hydrolysis WS)(NH4)2CO NH4OH and HCO3-weak base and weak acid ionKb for NH4OH = 1.6 x 10 -5Ka2 for H2CO3 = xSo (NH4)2CO3 is basic!
43 [HCN] [OH-] [CN-] Calculations !!! Kb = Kw Ka 1 x10-14 5.8 x 10-10 Calculate the pH of a M solution of KCN. ( Ka for HCN is 5.8 x 10-10)Salt is abasic salt,so need touse Kb….CN- + H2O HCN + OH-0.5M x x- ionmakesOH- ![HCN] [OH-][CN-]Kb =KwKa1 x10-145.8 x 10-10Kb =1.7 x10-5==
44 x20.5 –x1.7 x10-5=/x = 2.9 x 10-3 MpOH = -log(2.9 x 10-3) == = pH
46 HA(aq) + H2O(l) H3O+(aq) + A-(aq) BuffersSolutions that resist pH change when small amounts of acid or base are added.Two typesweak acid and its saltweak base and its saltHA(aq) + H2O(l) H3O+(aq) + A-(aq)Add OH Add H3O+shift to right shift to leftBased on the common ion effect.AttacksHA
47 Buffers solution with this organization: HC2H3O2 H+ + C2H3O2- A solution of a weak acid and its’ salt!Acetic Acid and Sodium Acetate:HC2H3O NaC2H3O2Mix these two compounds in water gives asolution with this organization:HC2H3O2 H C2H3O2-Concentration of the C2H3O2- would be greaterthan just a solution of low soluble HC2H3O2 !!!!
49 How does a buffer work? Ethanoic acid and ethanoate: Both: CH3CO2 - AND CH3CO2H are present in solutionCH3CO H3O+ HCH3CO2 + H2OThe ethanoate ions will be able to absorb anyexcess acid that is addedHCH3CO2 + OH- CH3CO H2OThe ethanoic acid will be able to absorb anyexcess base that is added
50 BuffersHenderson-Hasselbalch equationThe pH of a buffer does not depend on the absolute amount of the conjugate acid-base pair. It is based on the ratio of the two.Calculating pH of an acidic buffer use:pH = pKa + logCalculating the pH of a basic buffer use:pH = 14 - ( pKb + log )[A-][HA][HA][A-]
51 Buffers The beauty of a buffer…. A total of 100 ml of a 1.0 M HCl is add in 10 ml increments to 100 ml of each of the following:Solution A: Pure water - pH = 7(no acid in it ! )Solution B:A solution containing1.0 M HA (acid in it ),1.0 M A- (conjugate base in it )with a pKa of 7.0Calculate the pH after each addition.
52 Buffers Initially, each sample is at pH 7.00 After adding 10 ml of 1.0 M HCl we have:Pure water[H3O+] = =pH = 1.04From pH 7 to pH 1.04…..This is a pretty big jump!(10 ml)(1.0 M)(110 ml)
53 Buffers CH3CO2- + H3O+ HCH3CO2 + H2O Add 10 ml of 1.0 M HCl to our buffered system..10 moles moles moles- .01 moles molesmolesWe started with 0.10 moles of both the acid and conjugate base forms.The HCl in our first 10 ml can be expected to react with the conjugate base, converting it to the acid.After addition, we would have 0.09 moles of the base form and 0.11 moles of the acid form.CH3CO H3O+ HCH3CO2 + H2O
54 Buffers Checking the pH: Since all we need to be concerned about is the ratio of A- to HA in the 10-90% region then:pH = pKA + log= log= 6.91A change of only 0.09 (Buffer solution)compared to 5.96 ( in pure water) ![A-][HA]0.09 mol0.11 mol
55 Buffers Matter of fact…….. ml HCl pH added unbuffered buffered
58 IndicatorsAcid-base indicators are highly colored weak acids or bases.HIndic Indic-color color 2They may have more than one color transition.Example. Thymol blueRed Yellow BlueOne of the forms may be colorless - phenolphthalein (colorless to pink)
59 Indicators Indicator color change. Indicators are commonly used to detect the endpoint of an acid-base titration.The indicator should not start to undergo a color change until about one pH unit after the equivalence point.
60 Indicators Selection of an indicator. Under these conditions, the transition range for our indicator is pKa + 1.Ideally then, you want your indicator pKa to be one above the pKa for an acid sample.For bases, it should be one lower.You can seldom find a ‘perfect’ indicator so should use one that is a maximum of one unit away.
61 Indicators pH transition Indicator range color Bromophenol Blue yellow-blueMethyl Orange red-yellowMethyl Red red-yellowBromothymol Blue yellow-blueCresol Purple yellow-purplePhenolphthalein colorless - pinkThymolphthaleine colorless - blueAlizerin Yellow GG yellow - red
62 Indicator examplesAcid-base indicators are weak acids that undergo a color change at a known pH.pHphenolphthalein
64 Titrations revisited Methods based on measurement of volume. If the concentration of an acid is known, the amount of a base can be found.If we know the concentration of the base, then we can determine the amount of acid.All that is needed is some calibrated glassware and either an indicator or pH meter.
65 Titrations Buret - volumetric glassware used for titrations. It allows you to add a known amount of your titrant to the solution you are testing.If a pH meter is used, the equivalence point can be measured.An indicator will give you the endpoint.
66 TitrationsNote the color change which indicates that the ‘endpoint’ has been reached.StartEnd
67 Titration curves Acid-base titration curve A plot of the pH against the amount of acid or base added during a titration.Plots of this type are useful for visualizing a titration.It also can be used to show where an indicator undergoes its color change.
68 Titration curves Equivalence pH Point % titration or ml titrant Four regions of titration curveOvertitrationIndicatorTransitionEquivalencePointpHBuffer region% titration or ml titrant
69 Titration curves Strong acid titrated with a strong base. This is pretty straight forward since the net reaction is:H3O+ + OH H2OPrior to equivalence pointThe pH indicates the amount of sample present after accounting for dilution.[H3O+] =moles acid - moles basetotal volume in liters
70 Titration curves Strong acid titrated with a strong base. Equivalence point[H3O+] = [OH-]pKW = 14 = pH + pOHpH = 7.00So the equivalence point for strong acid/base problems is always at pH=7.00
71 Titration curves Strong acid titrated with a strong base. OvertitrationPast the equivalence point, we don’t have any acid remaining. All that we are doing is diluting our titrant by adding more solution volume and more OH- ions.[OH-] =pH = 14 - pOHmoles excesstotal volume in liters
72 Titration curves Example. Construct a titration curve for the titration of 100 ml of 0.10 M HCl with 0.10 M NaOHA. 0% titrationpH = -log[0.10] = 1B. After 10 ml NaOH[H3O+] == M, pH = 1.09(100 ml)(0.10 M) - (10 ml)(0.10 M)100 ml + 10 ml
73 Titration curves ml titrant total ml [H3O+] pH 0 100 0.10 1.00
76 Titration curves Reached the Equivalence point or 100 % titration (100ml acid)(0.10M acid) = (100ml base)( 0.10M base)pH MUST be x10-14 = (1 x10-7)2Note that for the first 90 ml of our titration, we only saw a change of 1.28 pH units.Now we have a jump of 4.72 pH units.
77 Titration curves Anything over the equilvalence point is Overtitration: (no more acid there!)Describing the dilution of our titrant:10 ml overtitration[OH-] = M= MpOH = 2.32pH = =(.10M)10 ml210 ml
78 Titration curves ml titrant total volume [OH-] pH 110 210 0.0048 11.68 Continue the calculations:ml titrant total volume [OH-] pH
79 Titration curvesJust adding more OH-Over titrationpHml NaOH
80 Titration curves Titration of a strong base with a strong acid. This is not significantly different from our earlier example.If you plot pOH rather that pH, the results would look identical.Typically we still plot pH versus ml titrant so the curves are inverted.
81 Titration curves basic sample (base in the beaker) pH acidic sample (acid in the beaker)ml titrant
82 Titration of weak acids or bases Titration of a weak acid or base with a strong titrant is a bit more complex than the strong acid/strong base example.We must be concerned with conjugate acid/base pairs and their equilibria.ExampleHA + H2O H3O+ + A-acidbase
83 Titration of weak acids or bases First, we’ll only be concerned about the titration of a weak acid with a strong base or a weak base with a strong acid.We still have the same four general regions for our titration curve.The calculation will require that you use the appropriate Ka or Kb relationship.We’ll start by reviewing the type of calculations involved and then work through an example.
84 Titration of weak acids or bases If your sample is an acid then useKa =At this point [H3O+] = [A-].You can solve for [H3O+] by using either the quadratic or approximation approach.Finally, calculate the pH.[H3O+][A-][HA]
85 Titration of weak acids or bases If your sample is an base then useKb =At this point [OH-] = [HA].You can solve for [OH-] by using either the quadratic or approximation approach.Then determine pH as 14 - pOH.[OH-][HA][A-]
86 Titration of weak acids or bases In this region, the pH is a function of the K value and the ‘ratio’ of the acid and base forms of our answer.A common format for the equilibrium expression used in this region is the Henderson-Hasselbalch equation.We can present it in two forms depending on the type of material we started with.
87 Titration of weak acids or bases Starting with an acidpH = pKa + logStarting with a basepH = 14 - ( pKb + log )( We’re just determining the pOH and then converting it to pH. )[A-][HA][HA][A-]
88 Titration of weak acids or bases Another approach that can be taken is to simply use the % titration values.For an acidic sample, you would use:pH = pKa + log% titration100 - % titration
89 Titration of weak acids or bases These equations have their limits and may break down if:Ka or Kb > 10-3You are working with very dilute solutions.In those cases, you need to consider the equilibrium for water
90 Titration of weak acids or bases 100% titration - the equivalence point.At the point, we have converted all of our sample to the conjugate form.If your sample was an acid, now solve for the pH using the Kb relationship -- do the opposite if you started with a base.Remember that pKa + pKb = 14You must also account for dilution of your sample as a result of adding titrant.
91 Titration of weak acids or bases Overtitration (> 100%)These calculations are identical to those covered in our strong acid/strong base example.You simply need to account for the amount of excess titrant added and how much it has been diluted.
92 Titration of weak acids or bases ExampleA 100 ml solution of 0.10 M benzoic acid is titrated with 0.10 M NaOH. Construct a titration curve.For benzoic acidKa = x 10-5pKa = 4.20
93 Titration of weak acids or bases Before adding any base:Ka =[H3O+] = [A-][HA] + [A-] = 0.10 M(We’ve assumed that [A-] is negligible compared to [HA].)[H3O+][A-][HA]
94 Titration of weak acids or bases Initial pH, before adding any base:Ka = x =x = (6.28 x )(0.10)= MpH = 2.60x20.10
95 Titration of weak acids or bases adding 10ml increments of base:Here we can use the Henderson-Hasselbalch equation in ml titration formatpH = pKa + logpH = log (10 / 90)= 3.25We can calculate other points by repeating this process.(increments of 10ml)(100ml – total added)(Showing the firstincrement of 10ml)
96 Titration of weak acids or bases ml added pHNote:At 50 ml titration,pH = pKaAlso, the was onlya change of 1.91pH units as wewent from 10 to90 ml titration.(5.15 – 3.24)
98 Titration of weak acids or bases 100% titration (EQUILVALENCE POINT)At this point, virtually all of our acid has been converted to the conjugate base - benzoate.Need to use the Kb relationship to solve for this point! (have no acid left!)Kb =Kb = Kw / Ka = x 10-10[OH-] [HA][A-]
99 Titration of weak acids or bases EQUILVALENCE POINTAt the equivalence point:[HA] = [OH-](We’ve diluted the[A-] = 0.05 M sample and the total volume at this point is 200 ml)Finally, assume that [benzoic acid] is negligible compared to [benzoate]..10m.200 liters
100 Titration of weak acids or bases 100% titration ( EQUILVALENCE POINT)Kb = x =x = (1.59 x )(0.050)= x 10-6 MpOH = pH = 14 - pOH = 8.45x20.050
101 Titration of weak acids or bases pH% titration
102 Titration of weak acids or bases OvertitrationAll we need to do here is to account for the dilution of our titrant.10 % overtitration (10 ml excess)[OH-] = M= MpOH = 2.32pH = =
103 Titration of weak acids or bases ml titrant total volume [OH-] pHThis is identical to what we obtained for our strong acid/strong base example
105 Polyprotic acidsA number of acids exist that have more that one ionizable hydrogen.ExamplesH3PO4 Phosphoric AcidH2SO4 Sulfuric AcidH2C2O4 Oxalic acidH2CO3 Carbonic acidEach H+ will ionize with its own constant. Also, it becomes more difficult to remove subsequent H+.
106 Stepwise equilibrium Example - H3PO4 Each hydrogen is capable of dissociation but not to the same extent at any given pH.H3PO H2PO HPO PO43-Since the removal of any one H+ affects how easily subsequent H+ are removed, each step has its own Ka value.Ka1Ka2Ka3
107 Stepwise equilibrium Note: [H3O+] is the same for each expression. The relative amountsof each species canbe found if the pHis known.The actual amountscan be found if pHand total H3PO4 isknown.[H3O+] [H2PO4-][H3PO4]KA1 =[H3O+] [HPO42-][H2PO4-]KA2 =[H3O+] [PO43-][HPO42-]KA3 =